RBSE Solutions for Class 8 Maths Chapter 6 Polygons Ex 6.2 is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 6 Polygons Exercise 6.2.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 6 |

Chapter Name |
Polygons |

Exercise |
Exercise 6.2 |

Number of Questions |
8 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 8 Maths Chapter 6 Polygons Ex 6.2

Question 1.

Fill in the blanks choosing the right option.

(i) Adjacent angles of a parallelogram are ……… (equal/supplement)

(ii) Diagonals of a rectangle are ……… (equal/perpendicular bisect)

(iii) In any trapezium AB || CD, If A = 100° then the value of ∠D will be ……(100/80°)

(iv) If in any quadrilateral, diagonal bisect each other on right angle then, it is called ……(parallelogram/rhombus)

(v) All squares are……(congruent/similar)

Solution

(i) supplementary

(ii) equal

(iii) 80°

(iv) rhombus

(v) similar.

Question 2.

In figure, BEST is a parallelogram. Find the values of x, y and z.

Solution

∵ BEST is a parallelogram.

∴ ∠B + x° = 180°

Adjacent angles of a parallelogram are supplementary

⇒ 110° + x° = 180°

⇒ x° = 180° – 110°

⇒ x° = 70°

⇒ x = 70

∵ Opposite angles of a parallelogram are equal

∴ x° = y°

⇒ 70° = y°

⇒ y = 70

and z° = 110°

⇒ z = 110

Question 3.

In following parallelograms, find the unknown value of x, y, z.

Solution

(i) x° = 90°

=> x = 90

∆OAD in,

OA = OD | given that

∴ ∠OAD = ∠ODA (= y°)

and also we take

∠AOD + ∠OAD + ∠ODA = 180°

∵ Sum of angles of a triangle is 180°

⇒ 90° + y° + y° = 180°

⇒ 90° + 2y° = 180°

⇒ 2y° = 180° – 90° = 90°

⇒ y° = \(\frac { 90 }{ 2 }\) = 45°

⇒ y = 45

∵ diagonals of a parallelogram bisect to each other.

∴ OA = OC

∵ OA = OD

∴ OC = OD

∴ ∠DOC = 90°

∴ As above

z° = 45°

⇒ z = 45

(ii) y° = 112° ∵ Opposite angles of a parallelogram are equal

⇒ y = 112

40° + z° + y° = 180°

∵ Adjacent angles of a parallelogram are supplementary

⇒ 40° + z° + 112° = 180°

⇒ z° + 152° = 180°

⇒ z° = 180° – 152° = 28°

⇒ z = 28

In end,

x° + y° + 40° = 180°

∵ Sum of angles of a triangle is 180°

⇒ x° + 112° + 40° = 180°

⇒ x° + 152° = 180°

⇒ x° = 180° – 152°

⇒ x° = 28°

⇒ x = 28

Question 4.

In any parallelogram, the ratio of two adjacent angles is 1 : 5. Find the value of all angles of parallelogram.

Solution

Let ABCD is a parallelogram According to question

∠A : ∠B = 1 : 5

∠A + ∠B = 180°

∵Sum of two adjacent angles of a parallelogram is 180°

Sum of Ratio = 1 + 5 = 6

∴∠A = \(\frac { 1 }{ 6 }\) x 180° = 30°

∠B = \(\frac { 5 }{ 6 }\) x 80° = 150°

∵ Opposite angles of a parallelogram are equal

∠C = ∠A = 30°

and ∠D = ∠B = 150°

∴Angles of a parallelogram are 30°, 150°, 30° and 150°.

Question 5.

In the following figures RISK and STEW are parallelograms, find the values of x and y (length in cm)

Solution

(i) ∵ Opposite sides of a parallelogram are equal.

∴3x = 18

and 3y – 1 = 26

⇒ x = \(\frac { 18 }{ 3 }\) = 6

and 3y = 26 + 1

⇒ 3y = 27

⇒ y = \(\frac { 27 }{ 2 }\) = 9

So, x = 6 cm and y = 9 cm

(ii) ∵ Diagonals of a parallelogram bisect each other

x + y = 16….(1)

and y + 7 = 20….(2)

From equation (2),

y = 20 – 7 = 13

We put y = 13 in equation (1)

x + 13 = 16

=> x = 16 – 13 = 3

So, x = 3 cm and y = 13 cm

Question 6.

HOPE is a rectangle.Its diagonals intersect each other at S.Find the value of x if SH = 2x + 4 and SE = 3x + 1.

Solution

∵ Diagonal of a rectangle bisect each other

∴ SE = SO

and SH = SP

∴ EO = 2SE

= 2 (3x + 1)

and HP = 2 SH = 2 (2x + 4)

∵ Diagonal of a rectangle are equal.

∴ EO = HP

⇒ 2(3x + 1) = 2(2x + 4)

⇒ 3x + 1 = 2x + 4

⇒ 3x – 2x = 4 – 1

⇒ x = 3

Question 7.

PEAR is a rhombus. Find the value of x, y and z, also write the causes.

Solution

∵ Diagonals of a rhombus bisect each other.

∴ OA = OP

⇒ x = 5 cm. ∵ OP = 5 cm. (given that)

and OE = OR

⇒ y = 12 cm. ∵ OR = 12 cm. (given that)

∵ sides of a rhombus are equal.

∴ EP = PR

⇒ z = 13 cm.

Question 8.

In a trapezium PQRS, PQ || SR, find the value of ∠x and ∠y.

Solution

∵ PQ || SR

∴ x + 90° = 180°

⇒ x = 180° – 90° = 90°

Again ∴ PQRS is a quadrilateral

∴ x + y + 130° + 90° = 360°

∵ We know that sum of all angles of a quadrilateral = 360°

⇒ 90° + y + 130° + 90° = 360°

⇒ y + 310° = 360°

⇒ y = 360° – 310°

⇒ y = 50°

We hope the given RBSE Solutions for Class 8 Maths Chapter 6 Polygons Ex 6.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 6 Polygons Exercise 6.2, drop a comment below and we will get back to you at the earliest.

## Leave a Reply